Question

A parallelogram has sides of lengths 7 and 4, and one angle is to Pind the length of the diagonalt (Round your answers to the deceatces Enter your owers as a cooperated it! Need Help?

Answer 1

The length of the diagonal of the parallelogram is approximately √65 units.

Step-by-step explanation:

In a parallelogram with sides of length 7 and 4, and given that one angle is θ, we can find the length of the diagonal using the Law of Cosines. The Law of Cosines states that for any triangle ABC with sides a, b, and c and angle θ opposite side c:

c² = a² + b² - 2ab cos(θ)

In the given parallelogram, let a = 7, b = 4, and \( \theta \) be the angle opposite side c, which is the diagonal. Rearranging the formula to solve for c:

c = √a² + b² - 2ab cos(θ)

Substituting the values, we get:

c = √7² + 4² - 2 × 7 × 4 cos(θ)

Since the angle θ is not provided, we use the fact that the diagonals of a parallelogram bisect each other. Thus, cos(θ) = -1 and:

c = √7² + 4² + 2 × 7 × 4 = √49 + 16 + 56 = √121 = √11² = 11

So, the length of the diagonal is 11 units.

Answer 2

The length of the diagonal of the parallelogram is approximately 1.30 units.

Given a parallelogram with side lengths of 7 and 4, and one angle is
`\((\pi)/(6)\)` radians:

To find the length of the diagonal, we can use the cosine rule for parallelograms:

The diagonal length d can be found using the formula:

`\[ d = √(a^2 + b^2 - 2ab \cos(\theta)) \]`

Where:

- a and b are the side lengths of the parallelogram.

-
`\(\theta\)` is the angle between the given sides.

Given:
`\(a = 7\), \(b = 4\), and \(\theta = (\pi)/(6)\).`

Let's plug these values into the formula:

`\[ d = \sqrt{7^2 + 4^2 - 2 * 7 * 4 * \cos\left((\pi)/(6)\right)} \]`


`\[ d = \sqrt{49 + 16 - 56 * (√(3))/(2)} \]`

`\[ d = \sqrt{65 - 28√(3)} \]`

Calculating the value:

`\[ d \approx \sqrt{65 - 28√(3)} \approx 1.30 \] (rounded to two decimal places)`